Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $t = \dfrac{-5z^2 + 35z + 150}{2z^2 - 6z - 140} \div \dfrac{z - 10}{4z + 28} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{-5z^2 + 35z + 150}{2z^2 - 6z - 140} \times \dfrac{4z + 28}{z - 10} $ First factor out any common factors. $t = \dfrac{-5(z^2 - 7z - 30)}{2(z^2 - 3z - 70)} \times \dfrac{4(z + 7)}{z - 10} $ Then factor the quadratic expressions. $t = \dfrac {-5(z - 10)(z + 3)} {2(z - 10)(z + 7)} \times \dfrac {4(z + 7)} {z - 10} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac { -5(z - 10)(z + 3) \times 4(z + 7)} { 2(z - 10)(z + 7) \times (z - 10)} $ $t = \dfrac {-20(z - 10)(z + 3)(z + 7)} {2(z - 10)(z + 7)(z - 10)} $ Notice that $(z - 10)$ and $(z + 7)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {-20\cancel{(z - 10)}(z + 3)(z + 7)} {2\cancel{(z - 10)}(z + 7)(z - 10)} $ We are dividing by $z - 10$ , so $z - 10 \neq 0$ Therefore, $z \neq 10$ $t = \dfrac {-20\cancel{(z - 10)}(z + 3)\cancel{(z + 7)}} {2\cancel{(z - 10)}\cancel{(z + 7)}(z - 10)} $ We are dividing by $z + 7$ , so $z + 7 \neq 0$ Therefore, $z \neq -7$ $t = \dfrac {-20(z + 3)} {2(z - 10)} $ $ t = \dfrac{-10(z + 3)}{z - 10}; z \neq 10; z \neq -7 $